Answer:
The total surface are of the bowl is given by: 0.0532*pi m² (approximately 0.166533 m²)
Explanation:
The total surface area of the semi-spherical bowl can be decomposed in three different sections: 1) an outer semi-sphere of radius 12 cm, 2) an inner semi-sphere of radius 10 cm, and 3) the edge, which is a 2-dimensional ring with internal radius of 10 cm and external radius of 12 cm. We will compute the areas independently and then sum them all.
a) Outer semi-sphere:
A1 = 2*pi*r² = 2*pi*(12 cm)² = 288*pi cm² = 904.78 cm²
b) Inner semi-sphere:
A2 = 2*pi*(10 cm)² = 200*pi cm² = 628.32 cm²
c) Edge (Ring):
A3 = pi*(r1² - r2²) = pi*((12 cm)²-(10 cm)²) = pi*(144-100) cm² = 44*pi cm² = 138.23 cm²
Therefore, the total surface area of the bowl is given by:
A = A1 + A2 + A3 = 288*pi cm² + 200*pi cm² + 44*pi cm² = 532*pi cm² (approximately 1665.33 cm²)
Changing units to m², as required in the problem, we get:
A = 532*pi cm² * (1 m² / 10, 000 cm²) = 0.0532*pi m² (approximately 0.166533 m²)
I think this one's B. energy and work are both measured in joules.
Answer:
True or False
Explanation:
Because.....
easy 50% chance you are right
The answer is noble gas. Since noble gas are constant and
unreactive. They can still shape compounds with other elements.
Group 15 is also group 5A and Group 17 is also group 7A. Elements in these sets
do not typically form ionic bonds; they are more on creating covalent bonds
since they're non-metals.
Therefore, that leaves us with B. from Group 1. They are metals (but Hydrogen)
which respond violently with water, and they form ionic bonds, for they drop
outer electrons easily.
Explanation:
W = PE
W = mgh
1500 J = (20 kg) (9.8 m/s²) h
h = 7.65 m
Round as needed.
